### Video Transcript

By completing the table of values for π of π₯ equals π₯ plus two squared minus four, identify the correct graph of the quadratic function on the domain the closed interval of negative four to zero.

There are four additional parts of this question weβll consider once weβve chosen the correct graph. Weβve been instructed to complete the table of values for our function π of π₯ equals π₯ plus two squared minus four. To complete the table, weβll plug in values from the table into π₯ and solve for π of π₯. Starting with negative four, we have negative four plus two squared minus four. Negative two squared minus four would equal four minus four. Therefore, π of negative four equals zero. And weβll add that to our table. π of negative three equals negative three plus two squared minus four. Negative three plus two is negative one. Negative one squared is one. And one minus four equals negative three. π of negative three equals negative three.

When we plug in negative two, we find that π of negative two equals negative four, π of negative one equals negative three, and the last value in our table π of zero is equal to zero. Letβs use the points negative four, zero and zero, zero to try and narrow down the graphs. The functions in graph (A), (C), and (E) do not cross the points zero, zero and negative four, zero, which means we now need to consider the functions graphed in (B) and (D). To do that, we could choose another point. Letβs choose negative two, negative four. When we graph negative two, negative four, we see that this point does not fall on the function graphed in (B). Therefore, the correct graph is the graph in (D). Now weβre ready to consider the rest of the question.

Identify the π₯-intercepts of the given quadratic function. Identify the π¦-intercepts of the given quadratic function. Identify the axis of symmetry of the given quadratic function. And identify the vertex of the given quadratic function.

We know that the π₯-intercepts are the places where the graph crosses the π₯-axis. There are a few methods for finding the π₯-intercepts. However, as weβve already filled in the table, we know that our graph crosses the π₯-axis when π₯ equals negative four and when π₯ equals zero. Itβs worth noting here that weβre dealing with the domain that has a closed interval, and that means negative four and zero are both included in the interval. Therefore, the first π₯-intercept is at negative four and the second is at zero.

Now we need to identify the π¦-intercept of the given quadratic function. The π¦-intercept of a quadratic function is the place where the graph crosses the π¦-axis. Itβs the place where π₯ equals zero. Weβve calculated this point in our table, and we see it in the graph. This function crosses the π¦-axis at the point zero, zero. Therefore, the π¦-intercept equals zero.

Moving on, weβll identify the axis of symmetry. To do that, letβs consider what form weβve been given our quadratic function in. We know that π of π₯ equals π₯ plus two squared minus four. This is in the form π of π₯ equals π times π₯ minus β squared plus π. We call this vertex form. And for a function in vertex form, the axis of symmetry is located where π₯ equals β. Since the main form is π₯ minus β and we have π₯ plus two, we have to identify β as negative two. Therefore, the axis of symmetry is located at negative two. This checks out visually as this is the vertical line about which the parabola is symmetrical.

And finally, we need to calculate the vertex. In vertex form of a quadratic function, the vertex is located at the point β, π. Weβve identified β as negative two. In vertex form, you have plus π. Since weβre subtracting four, our π-value is then negative four, making the vertex negative two, negative four. Again, this checks out visually; since this parabola opens upward, the vertex is the minimum value. For the function π of π₯ equals π₯ plus two squared minus four, the π₯-intercepts are negative four, zero, the π¦-intercept is located at π¦ equals zero, the axis of symmetry π₯ equals negative two, and the vertex negative two, negative four.